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The value of ΔH° for the reaction below is -336 kJ. Calculate the heat (kJ) released to the surroundings when 23.0 g of HCl is formed. CH4(g) + 3Cl2(g) → CHCl3(l) + 3HCl(g)

Respuesta :

ardni313

The heat (kJ) released to the surroundings : -70.56 kJ

Further explanation

Reaction

CH₄(g) + 3Cl₂(g) → CHCl₃(l) + 3HCl(g) ΔH°=-336 kJ

For 3 mol HCl formed ⇒ ΔH°=-336 kJ

23.0 g of HCl :

  • mol

[tex]\tt \dfrac{23}{36.5}=0.63[/tex]

Heat released for 0.63 mol HCl :

[tex]\tt \dfrac{0.63}{3}\times -336=-70.56~kJ[/tex]

The heat (kJ) released to the surroundings at the given equation is 70.7KJ.

  • The calculation is as follows:

We know that

[tex]Moles\ or\ HCl\ formed = mass \div molar\ mass\\\\= 23.0\div 36.46[/tex]

= 0.631 mole

Since energy should be related at the time when 3 moles of HCI should be generated = 336 K

Now energy released is  

[tex]= (336\div 3)\times 0.631[/tex]

= 70.65 kJ

Hence we can conclude that heat released to the surrounding is 70.7 kJ.

Learn more:  brainly.com/question/17127206

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